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lesya692 [45]
3 years ago
11

Heat is added to a 200.-gram sample of H2O(s) to melt the sample at 0°C. Then the resulting H2O (image) is heated to a final tem

perature of 65°C. Determine the total amount of heat required to completely melt the sample.
Chemistry
2 answers:
pochemuha3 years ago
3 0
You are given 200 grams of H2O(s) at an initial temperature of 0°C. you are also given the final temperature of water after heating at 65°C. You are required to get the total amount of heat to melt the sample. The specific heat capacity, cp, of water is 4.186 J/g-°C. Let us say that T1 = 0°C and T2 = 65°C. The equation for heat, Q, is  

Q = m(cp)(T2-T1)
Q = 200g(4.186 J/g-°C )(65°C - 0°C)
<u>Q = 54,418J</u>
MaRussiya [10]3 years ago
3 0

Answer: 54.418 kJ amount of heat required to completely melt the sample.

Explanation:

Mass of water = 200 g

Initial temperature of the water = 0°C

Final  temperature of the water = 65°C

Temperature change =\Delta T=65^oC-0^oC=65^oC

Specific heat capacity = 4.186J/g^oC

Q=mC\Delta T=200 g\times 4.186J/g^oC\times 65^oC

Q=54,418 J=54.418 kJ  (1 kJ = 1000J)

54.418 kJ amount of heat required to completely melt the sample.

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Explanation:

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Hence, when it chemically combines another nitrogen atom then as both the atoms are non-metals. So, sharing of electrons will take place.

Also, there is no difference in electronegativity of two nitrogen atoms. Hence, compound formed N_{2} is non-polar covalent in nature.

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What is the type of mutation represented by the amino acid sequence below?
Ilia_Sergeevich [38]

Answer:

Substitution mutation

Explanation:

A substitution mutation is a type of mutation in which one or more nucleotide base is replaced by another in a sequence. This will result in the replacement of one or more amino acid in the amino acid sequence.

This is the case in this question where the original amino acid sequence was given as: Leucine – Alanine – Glycine – Leucine. After mutation, the following mutated sequence was produced: Leucine – Alanine – Valine – Leucine.

As illustrated above, one would notice that there is replacement of GLYCINE amino acid by VALINE in the mutated sequence, hence, it is an example of SUBSTITUTION MUTATION.

6 0
3 years ago
What is the absorbance of the permanganate in the diluted waste solution?
Snowcat [4.5K]
<span>Not to be confused with tetration. This article is about volumetric titration. For other uses, see Titration (disambiguation). Acid–base titration is a quantitative analysis of concentration of an unknown acid or base solution. Titration, also known as titrimetry,[1] is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator[2] is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand[3] to determine concentration. The volume of titrant reacted is called titration volume</span>
7 0
3 years ago
a solution of sodium hydroxide, 0.0500 M, is used to titrate a 15.00 mL sample of hydrochloric acid to the endpoint. The initial
Akimi4 [234]

Answer:

The answer to your question is Molarity = 0.0708

Explanation:

Data

NaOH  0.05 M     Volume 1 = 3.87 ml    Volume 2 = 25.11 ml

HCl   15 ml

Process

1.- Find the volume used of NaOH

                              25.11 - 3.87 = 21.24 ml = 0.02124 l

2.- Write the balanced equation of the reaction

                   NaOH  +  HCl   ⇒   NaCl + H₂O

3.- Calculate the moles of NaOH in the solution

Molarity = \frac{moles}{volume}

moles = Molarity x volume

moles = 0.05 x 0.02124

moles = 0.001062

4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.

                   1 mol of NaOH -------------  1 mol of HCl

 0.001062 moles of NaOH ------------    x

                  x = (0.001062 x 1) / 1

                  x = 0.001062 moles of HCl

5.- Find the molarity of HCl

Molarity = \frac{0.001062}{0.015}

Molarity = 0.0708

4 0
3 years ago
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

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3 years ago
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